My approach to solve this questions has been dividing the problem in 2 parts.
First Part.
The numbers that started with 37, since 22, 33, 31, etc. are not really options. So in the first two places I only have 1 possibility (37), leaving 1,2,2,3,3,7 available. So the numbers of combinations for this are:
$1\times5\times4\times3\times2\times1=240$
Second Part.
The numbers that started with 7, leaving 1,2,2,3,3,3 available. So, the number of combinations for this are:
$1\times6\times5\times4\times3\times2\times1=720$
Therefore the numbers than can be formed are: $240+720=960$
Any help is greatly appreciated.
You have to get rid of multiplicity as some digits are not distinguishable.
For the first case, choosing from $1,2,2,3,3$ should be
$$\frac{5!}{2!2!}$$
For the second case, choosing from $1,2,2,3,3,3$ should be
$$\frac{6!}{2!3!}$$
$$\frac{5!}{2!2!}+\frac{6!}{2!3!}=\frac{5!}{2!2!}\left(1+ \frac{6}{3}\right)=\frac{360}{4}=90$$